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05 Jan 2022, 04:00
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
32% (01:37) correct
68%
(01:45)
wrong
based on 93
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If a and b are two distinct positive integers, what is the remainder when a divides b?
(1) When b is divided by the lowest number that is divisible by both a and b, the result is an integer.
(2) a and b have the same prime factors.
(1) When b is divided by the lowest number that is divisible by both a and b, the result is an integer.
(2) a and b have the same prime factors.
18 Jan 2022, 22:16
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Bunuel
Subtleties.! Damn..!
Points that will come in handy:
1) a and b are two distinct positive integers
2) a is dividing b
We need to find the remainder.
(1) When b is divided by the lowest number that is divisible by both a and b, the result is an integer.
What is the lowest number that is divisible by both a and b.?
It's the LCM.
So when b is divided by the LCM the result is an integer.
Now let's plug in values to make our brain understand it better:
Let a be 2 and b be 4.
So, LCM will be 4
So when 4 is divided by 4 we are having 1(quotient) as an Integer.
This means that b is the bigger integer because a divided by LCM won't fetch us an Integer value.
So remainder when a divides b will be 0.
Let's take another value:
2 and 3
LCM 6
But here b is not divisible by LCM, so we can't take this equation.
Basically, the value of the LCM should be equal to the value of the bigger integer.
Sufficient.
(2) a and b have the same prime factors.
a can be 2 b can 4 in this case the remainder is zero.
but when a = 4 and b = 2 in this case the remainder is 2.
No unique answer is possible.
Not Sufficient.
A is the answer.
18 Jan 2022, 23:54
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Ouch, talk about not reading the question stem well.
“When a divides b”
B/A ———> I, of course, read the question the opposite way.
S1: the lowest number that is divisible by two distinct positive integers A and B = Lowest Common Multiple of A and B
The only way B, itself, could be divisible by the LCM of B and another integer is when:
B = LCM(A and B) ——-> in which case the integer result is 1
In other words, statement 1 is effectively telling us that B is the LCM of (A and B)
If B = LCM(A and B)
———- and from the definition of the LCM, we know that the LCM is the lowest integer into which A and B will divide
——— A will divide evenly into the LCM(A and B)
—- which means A will divide evenly into B
No remainder. Statement 1 is sufficient
Statement 2 can be proven insufficient by using two different cases in which A and B have the same prime factors.
B = 2 and A = 4 ——- remainder of 2
B = 3 and A = 9 ——- remainder of 3
A
s1 alone sufficient.
Great question
Posted from my mobile device
“When a divides b”
B/A ———> I, of course, read the question the opposite way.
S1: the lowest number that is divisible by two distinct positive integers A and B = Lowest Common Multiple of A and B
The only way B, itself, could be divisible by the LCM of B and another integer is when:
B = LCM(A and B) ——-> in which case the integer result is 1
In other words, statement 1 is effectively telling us that B is the LCM of (A and B)
If B = LCM(A and B)
———- and from the definition of the LCM, we know that the LCM is the lowest integer into which A and B will divide
——— A will divide evenly into the LCM(A and B)
—- which means A will divide evenly into B
No remainder. Statement 1 is sufficient
Statement 2 can be proven insufficient by using two different cases in which A and B have the same prime factors.
B = 2 and A = 4 ——- remainder of 2
B = 3 and A = 9 ——- remainder of 3
A
s1 alone sufficient.
Great question
Bunuel
Posted from my mobile device
